Re: [algogeeks] Re: Array Merge Problem
@ross: I couldn't get reddy's solution. Please explain.
On Sun, Jun 5, 2011 at 10:50 PM, Deepak Jha deepak.127.0@gmail.comwrote:
the below solution should work given the input array is sorted ( I am
assuming ascending order)
void rearrangeArray(int[] a, int[] b){
int m = a.length;
int n = b.length;
int i = m - 1;
int j = 0;
while((i =0) (j = n-1)){
if(a[i] b[j]){
int temp = a[i];
a[i] = b[j];
b[j] = temp;
}
i--;
j++;
}
}
On Sat, Jun 4, 2011 at 2:29 PM, ross jagadish1...@gmail.com wrote:
Hi Rohit all,
Sorry that there was a small typo in the 'n' 'm' texts.
The example given by me is anyway the correct one.
Sravan Reddy's solution worked fine.
On Jun 4, 10:08 am, rohit rajuljain...@gmail.com wrote:
i think solution would be like this
eg:
A : 1 2 3 B: 0 1.5 4 5 9
Output:
A can contain any combination of nos 0,1,1.5
and B should contain 2 3 4 5 9 (in any order.)
this example is given by ROSS itself.
so sravanreddy solution is right , correct me if i'm wrong.
On Jun 3, 8:07 pm, bittu shashank7andr...@gmail.com wrote:
@sravanreddy...logical bugs if A is size of n B is size m from your
example assuming nm so if you want smallest m elements in A then u
only capacity of n elements didn't allocate memory so these elements
initialized by INT_MIN for m-n nodes so thatarrayA can hold m
smallest elements then what r u swapping u dude..isn't garbage
value ?? you will get at 1st step only just run it ?? in you algo
A_End=m-1(which 4th position inArraythat DNE)..?? also you have to
free memory for m-n inarrayB as it contains n largest elements .
take
A= 1,2,3 n=3
B= 0,1,4,5,9 m=5
after allocating memory toArrayA for m-n elements A will looks
likes 1 2 3 INT_Max INT_Max
now what you wants A should contains m smallest elements B have n
largest elements
so O/P should be A=1,2,3,1,0 B=INT_Max,INT_Max,4,5,9 now free
memory used by 1st elements inarrayB so that A will represent M
smallest elements B will have n Largest elements
so that above will work.
Hope I am Correct let me know if any issue with explanation
Thanks
ShashankThe Best Way To Escape From Theproblemis To Solve It
CSE,BIT Mesra
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[algogeeks] Re: Array Merge Problem
@aakash Johari:
Let a and b be the 2 arrays.
At each stage of the process, if an element of A is greater than B,
then swap the largest element of A with the smallest element of B
and adjust pointers.
A : 2 4 15 12
B : 0.2 1 33 44
Now, 20, therefore swap 0 with 12..
Every step of the process, gets in the smallest elemnt of A and swaps
it
with the largest element of B.
Hope its clear.
On Jun 6, 11:15 am, Aakash Johari aakashj@gmail.com wrote:
@ross: I couldn't get reddy's solution. Please explain.
On Sun, Jun 5, 2011 at 10:50 PM, Deepak Jha deepak.127.0@gmail.comwrote:
the below solution should work given the input array is sorted ( I am
assuming ascending order)
void rearrangeArray(int[] a, int[] b){
int m = a.length;
int n = b.length;
int i = m - 1;
int j = 0;
while((i =0) (j = n-1)){
if(a[i] b[j]){
int temp = a[i];
a[i] = b[j];
b[j] = temp;
}
i--;
j++;
}
}
On Sat, Jun 4, 2011 at 2:29 PM, ross jagadish1...@gmail.com wrote:
Hi Rohit all,
Sorry that there was a small typo in the 'n' 'm' texts.
The example given by me is anyway the correct one.
Sravan Reddy's solution worked fine.
On Jun 4, 10:08 am, rohit rajuljain...@gmail.com wrote:
i think solution would be like this
eg:
A : 1 2 3 B: 0 1.5 4 5 9
Output:
A can contain any combination of nos 0,1,1.5
and B should contain 2 3 4 5 9 (in any order.)
this example is given by ROSS itself.
so sravanreddy solution is right , correct me if i'm wrong.
On Jun 3, 8:07 pm, bittu shashank7andr...@gmail.com wrote:
@sravanreddy...logical bugs if A is size of n B is size m from your
example assuming nm so if you want smallest m elements in A then u
only capacity of n elements didn't allocate memory so these elements
initialized by INT_MIN for m-n nodes so thatarrayA can hold m
smallest elements then what r u swapping u dude..isn't garbage
value ?? you will get at 1st step only just run it ?? in you algo
A_End=m-1(which 4th position inArraythat DNE)..?? also you have to
free memory for m-n inarrayB as it contains n largest elements .
take
A= 1,2,3 n=3
B= 0,1,4,5,9 m=5
after allocating memory toArrayA for m-n elements A will looks
likes 1 2 3 INT_Max INT_Max
now what you wants A should contains m smallest elements B have n
largest elements
so O/P should be A=1,2,3,1,0 B=INT_Max,INT_Max,4,5,9 now free
memory used by 1st elements inarrayB so that A will represent M
smallest elements B will have n Largest elements
so that above will work.
Hope I am Correct let me know if any issue with explanation
Thanks
ShashankThe Best Way To Escape From Theproblemis To Solve It
CSE,BIT Mesra
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Re: [algogeeks] Re: Array Merge Problem
the below solution should work given the input array is sorted ( I am
assuming ascending order)
void rearrangeArray(int[] a, int[] b){
int m = a.length;
int n = b.length;
int i = m - 1;
int j = 0;
while((i =0) (j = n-1)){
if(a[i] b[j]){
int temp = a[i];
a[i] = b[j];
b[j] = temp;
}
i--;
j++;
}
}
On Sat, Jun 4, 2011 at 2:29 PM, ross jagadish1...@gmail.com wrote:
Hi Rohit all,
Sorry that there was a small typo in the 'n' 'm' texts.
The example given by me is anyway the correct one.
Sravan Reddy's solution worked fine.
On Jun 4, 10:08 am, rohit rajuljain...@gmail.com wrote:
i think solution would be like this
eg:
A : 1 2 3 B: 0 1.5 4 5 9
Output:
A can contain any combination of nos 0,1,1.5
and B should contain 2 3 4 5 9 (in any order.)
this example is given by ROSS itself.
so sravanreddy solution is right , correct me if i'm wrong.
On Jun 3, 8:07 pm, bittu shashank7andr...@gmail.com wrote:
@sravanreddy...logical bugs if A is size of n B is size m from your
example assuming nm so if you want smallest m elements in A then u
only capacity of n elements didn't allocate memory so these elements
initialized by INT_MIN for m-n nodes so thatarrayA can hold m
smallest elements then what r u swapping u dude..isn't garbage
value ?? you will get at 1st step only just run it ?? in you algo
A_End=m-1(which 4th position inArraythat DNE)..?? also you have to
free memory for m-n inarrayB as it contains n largest elements .
take
A= 1,2,3 n=3
B= 0,1,4,5,9 m=5
after allocating memory toArrayA for m-n elements A will looks
likes 1 2 3 INT_Max INT_Max
now what you wants A should contains m smallest elements B have n
largest elements
so O/P should be A=1,2,3,1,0 B=INT_Max,INT_Max,4,5,9 now free
memory used by 1st elements inarrayB so that A will represent M
smallest elements B will have n Largest elements
so that above will work.
Hope I am Correct let me know if any issue with explanation
Thanks
ShashankThe Best Way To Escape From Theproblemis To Solve It
CSE,BIT Mesra
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[algogeeks] Re: Array Merge Problem
i think solution would be like this eg: A : 1 2 3 B: 0 1.5 4 5 9 Output: A can contain any combination of nos 0,1,1.5 and B should contain 2 3 4 5 9 (in any order.) this example is given by ROSS itself. so sravanreddy solution is right , correct me if i'm wrong. On Jun 3, 8:07 pm, bittu shashank7andr...@gmail.com wrote: @sravanreddy...logical bugs if A is size of n B is size m from your example assuming nm so if you want smallest m elements in A then u only capacity of n elements didn't allocate memory so these elements initialized by INT_MIN for m-n nodes so thatarrayA can hold m smallest elements then what r u swapping u dude..isn't garbage value ?? you will get at 1st step only just run it ?? in you algo A_End=m-1(which 4th position inArraythat DNE)..?? also you have to free memory for m-n inarrayB as it contains n largest elements . take A= 1,2,3 n=3 B= 0,1,4,5,9 m=5 after allocating memory toArrayA for m-n elements A will looks likes 1 2 3 INT_Max INT_Max now what you wants A should contains m smallest elements B have n largest elements so O/P should be A=1,2,3,1,0 B=INT_Max,INT_Max,4,5,9 now free memory used by 1st elements inarrayB so that A will represent M smallest elements B will have n Largest elements so that above will work. Hope I am Correct let me know if any issue with explanation Thanks ShashankThe Best Way To Escape From Theproblemis To Solve It CSE,BIT Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Array Merge Problem
Hi Rohit all, Sorry that there was a small typo in the 'n' 'm' texts. The example given by me is anyway the correct one. Sravan Reddy's solution worked fine. On Jun 4, 10:08 am, rohit rajuljain...@gmail.com wrote: i think solution would be like this eg: A : 1 2 3 B: 0 1.5 4 5 9 Output: A can contain any combination of nos 0,1,1.5 and B should contain 2 3 4 5 9 (in any order.) this example is given by ROSS itself. so sravanreddy solution is right , correct me if i'm wrong. On Jun 3, 8:07 pm, bittu shashank7andr...@gmail.com wrote: @sravanreddy...logical bugs if A is size of n B is size m from your example assuming nm so if you want smallest m elements in A then u only capacity of n elements didn't allocate memory so these elements initialized by INT_MIN for m-n nodes so thatarrayA can hold m smallest elements then what r u swapping u dude..isn't garbage value ?? you will get at 1st step only just run it ?? in you algo A_End=m-1(which 4th position inArraythat DNE)..?? also you have to free memory for m-n inarrayB as it contains n largest elements . take A= 1,2,3 n=3 B= 0,1,4,5,9 m=5 after allocating memory toArrayA for m-n elements A will looks likes 1 2 3 INT_Max INT_Max now what you wants A should contains m smallest elements B have n largest elements so O/P should be A=1,2,3,1,0 B=INT_Max,INT_Max,4,5,9 now free memory used by 1st elements inarrayB so that A will represent M smallest elements B will have n Largest elements so that above will work. Hope I am Correct let me know if any issue with explanation Thanks ShashankThe Best Way To Escape From Theproblemis To Solve It CSE,BIT Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Array Merge Problem
@sravanreddy...logical bugs if A is size of n B is size m from your example assuming nm so if you want smallest m elements in A then u only capacity of n elements didn't allocate memory so these elements initialized by INT_MIN for m-n nodes so that array A can hold m smallest elements then what r u swapping u dude..isn't garbage value ?? you will get at 1st step only just run it ?? in you algo A_End=m-1(which 4th position in Array that DNE)..?? also you have to free memory for m-n in array B as it contains n largest elements . take A= 1,2,3 n=3 B= 0,1,4,5,9 m=5 after allocating memory to Array A for m-n elements A will looks likes 1 2 3 INT_Max INT_Max now what you wants A should contains m smallest elements B have n largest elements so O/P should be A=1,2,3,1,0 B=INT_Max,INT_Max,4,5,9 now free memory used by 1st elements in array B so that A will represent M smallest elements B will have n Largest elements so that above will work. Hope I am Correct let me know if any issue with explanation Thanks ShashankThe Best Way To Escape From The problem is To Solve It CSE,BIT Mesra -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Array Merge Problem
Please try this solution. And tell if it fails at any case. If it works
fine, I will tell the logic.
#include stdio.h
#include stdlib.h
void merge(int *a, int m, int *b, int n)
{
int i, j, k;
int t;
i = j = 0;
k = -1;
while ( i m a[i] b[j] ) {
i++;
}
while ( i m j n ) {
if ( k == -1 b[j] a[i] ) {
t = a[i]; a[i] = b[j]; b[j] = t;
k = j;
j++;
i++;
} else if ( b[j] b[k] ) {
t = a[i]; a[i] = b[j]; b[j] = t;
j++;
i++;
} else {
t = a[i]; a[i] = b[k]; b[k] = t;
i++;
}
}
}
int main()
{
int m, n;
int *a, *b;
int i;
scanf (%d%d, m, n);
a = (int*) malloc(sizeof(int) * m);
b = (int*) malloc(sizeof(int) * n);
for ( i = 0; i m; i++ )
scanf (%d, a + i);
for ( i = 0; i n; i++ )
scanf (%d, b + i);
merge (a, m, b, n);
printf (After Merge Operation : \n);
printf (1st Array : );
for ( i = 0; i m; i++ ) {
printf (%d , a[i]);
}
printf (\n2nd Array : );
for ( i = 0; i n; i++ ) {
printf (%d , b[i]);
}
return 0;
}
On Fri, Jun 3, 2011 at 8:07 AM, bittu shashank7andr...@gmail.com wrote:
@sravanreddy...logical bugs if A is size of n B is size m from your
example assuming nm so if you want smallest m elements in A then u
only capacity of n elements didn't allocate memory so these elements
initialized by INT_MIN for m-n nodes so that array A can hold m
smallest elements then what r u swapping u dude..isn't garbage
value ?? you will get at 1st step only just run it ?? in you algo
A_End=m-1(which 4th position in Array that DNE)..?? also you have to
free memory for m-n in array B as it contains n largest elements .
take
A= 1,2,3 n=3
B= 0,1,4,5,9 m=5
after allocating memory to Array A for m-n elements A will looks
likes 1 2 3 INT_Max INT_Max
now what you wants A should contains m smallest elements B have n
largest elements
so O/P should be A=1,2,3,1,0 B=INT_Max,INT_Max,4,5,9 now free
memory used by 1st elements in array B so that A will represent M
smallest elements B will have n Largest elements
so that above will work.
Hope I am Correct let me know if any issue with explanation
Thanks
ShashankThe Best Way To Escape From The problem is To Solve It
CSE,BIT Mesra
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Re: [algogeeks] Re: Array Merge Problem
it can be done in O(m) . Use something like binary search .
code is here ...
#includestdio.h
void splitMN(int a[],int m , int b[], int n){
int al = 0 , bl = 0 ;
int ah = m-1 , bh = n-1 ;
int ai = (ah+al+1)/2;
int bi = (bh+bl+1)/2;
while(ai+bi!=m){
printf(Enter ai = %d, bi = %d\n ,ai,bi);
if(ai+bi m){
if(a[ai] b[bi]){
al = ai ;
if(al == ai) break;
ai = (ah+al+1)/2;
}else{
bl = bi ;
if(bh == bi) break;
bi = (bh+bl+1)/2;
}
}else{
if(a[ai] b[bi]){
ah = ai ;
if(ah == ai) break;
ai = (ah+al+1)/2;
}else{
bh = bi ;
if(bh == bi) break;
bi = (bh+bl+1)/2;
}
}
}
bi = 0 ;
ai ;
while(ai m){
a[ai] = a[ai]^b[bi]^(b[bi] = a[ai]);
ai++ ; bi++ ;
}
}
int main(){
int m , n ;
printf(Enter m , n : );
scanf(%d%d,m,n);
int a[m] , b[n] ;
int i ;
for(i = 0 ; i m ; i++)
scanf(%d,a[i]);
for(i = 0 ; i n ; i++)
scanf(%d,b[i]);
printf(Enter m , n : );
splitMN(a,m,b,n);
for(i = 0 ; i m ; i++)
printf(%d\t,a[i]);
printf(\n);
for(i = 0 ; i n ; i++)
printf(%d\t,b[i]);
printf(\n);
}
On Sat, Jun 4, 2011 at 4:03 AM, Aakash Johari aakashj@gmail.com wrote:
Please try this solution. And tell if it fails at any case. If it works
fine, I will tell the logic.
#include stdio.h
#include stdlib.h
void merge(int *a, int m, int *b, int n)
{
int i, j, k;
int t;
i = j = 0;
k = -1;
while ( i m a[i] b[j] ) {
i++;
}
while ( i m j n ) {
if ( k == -1 b[j] a[i] ) {
t = a[i]; a[i] = b[j]; b[j] = t;
k = j;
j++;
i++;
} else if ( b[j] b[k] ) {
t = a[i]; a[i] = b[j]; b[j] = t;
j++;
i++;
} else {
t = a[i]; a[i] = b[k]; b[k] = t;
i++;
}
}
}
int main()
{
int m, n;
int *a, *b;
int i;
scanf (%d%d, m, n);
a = (int*) malloc(sizeof(int) * m);
b = (int*) malloc(sizeof(int) * n);
for ( i = 0; i m; i++ )
scanf (%d, a + i);
for ( i = 0; i n; i++ )
scanf (%d, b + i);
merge (a, m, b, n);
printf (After Merge Operation : \n);
printf (1st Array : );
for ( i = 0; i m; i++ ) {
printf (%d , a[i]);
}
printf (\n2nd Array : );
for ( i = 0; i n; i++ ) {
printf (%d , b[i]);
}
return 0;
}
On Fri, Jun 3, 2011 at 8:07 AM, bittu shashank7andr...@gmail.com wrote:
@sravanreddy...logical bugs if A is size of n B is size m from your
example assuming nm so if you want smallest m elements in A then u
only capacity of n elements didn't allocate memory so these elements
initialized by INT_MIN for m-n nodes so that array A can hold m
smallest elements then what r u swapping u dude..isn't garbage
value ?? you will get at 1st step only just run it ?? in you algo
A_End=m-1(which 4th position in Array that DNE)..?? also you have to
free memory for m-n in array B as it contains n largest elements .
take
A= 1,2,3 n=3
B= 0,1,4,5,9 m=5
after allocating memory to Array A for m-n elements A will looks
likes 1 2 3 INT_Max INT_Max
now what you wants A should contains m smallest elements B have n
largest elements
so O/P should be A=1,2,3,1,0 B=INT_Max,INT_Max,4,5,9 now free
memory used by 1st elements in array B so that A will represent M
smallest elements B will have n Largest elements
so that above will work.
Hope I am Correct let me know if any issue with explanation
Thanks
ShashankThe Best Way To Escape From The problem is To Solve It
CSE,BIT Mesra
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(IIIT Allahabad)
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*With Regards :*
Ravinder Kumar
B.Tech 3rd Year
Computer Science and Engineering
MNNIT Allahabad
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[algogeeks] Re: Array Merge Problem
Maintain a pointer A_end = m-1;
doing a comparision something similar to merge sort
int i=0;j=0;
while (i m){
if (a[i] b[j])
i++;
else{
swap(a[A_end],b[j])
A_end --;
j++;
}
}
This runs in O(m) time and no extra space, also the sort order is not
guarenteed.
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[algogeeks] Re: Array Merge Problem
@sravanreddy:
Hey, Nice Solution :) cool!
On May 29, 7:44 am, sravanreddy001 sravanreddy...@gmail.com wrote:
Maintain a pointer A_end = m-1;
doing a comparision something similar to merge sort
int i=0;j=0;
while (i m){
if (a[i] b[j])
i++;
else{
swap(a[A_end],b[j])
A_end --;
j++;
}
}
This runs in O(m) time and no extra space, also the sort order is not
guarenteed.
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Re: [algogeeks] Re: Array Merge Problem
@sravanreddy: it won't work. Try 3,91,9 and 90,1,8,2,5 . correct me
if i m wrong.
Thanks,
Ankit Sambyal
On Sat, May 28, 2011 at 9:16 PM, ross jagadish1...@gmail.com wrote:
@sravanreddy:
Hey, Nice Solution :) cool!
On May 29, 7:44 am, sravanreddy001 sravanreddy...@gmail.com wrote:
Maintain a pointer A_end = m-1;
doing a comparision something similar to merge sort
int i=0;j=0;
while (i m){
if (a[i] b[j])
i++;
else{
swap(a[A_end],b[j])
A_end --;
j++;
}
}
This runs in O(m) time and no extra space, also the sort order is not
guarenteed.
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Re: [algogeeks] Re: Array Merge Problem
@Ankit, The input should be 2 sorted arrays
Thanks,
Immanuel
On Sun, May 29, 2011 at 10:48 AM, ankit sambyal ankitsamb...@gmail.comwrote:
@sravanreddy: it won't work. Try 3,91,9 and 90,1,8,2,5 . correct
me if i m wrong.
Thanks,
Ankit Sambyal
On Sat, May 28, 2011 at 9:16 PM, ross jagadish1...@gmail.com wrote:
@sravanreddy:
Hey, Nice Solution :) cool!
On May 29, 7:44 am, sravanreddy001 sravanreddy...@gmail.com wrote:
Maintain a pointer A_end = m-1;
doing a comparision something similar to merge sort
int i=0;j=0;
while (i m){
if (a[i] b[j])
i++;
else{
swap(a[A_end],b[j])
A_end --;
j++;
}
}
This runs in O(m) time and no extra space, also the sort order is not
guarenteed.
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Re: [algogeeks] Re: Array Merge Problem
Oh I didn't read the question properly, Thanks for pointing out...
On Sat, May 28, 2011 at 10:28 PM, immanuel kingston
kingston.imman...@gmail.com wrote:
@Ankit, The input should be 2 sorted arrays
Thanks,
Immanuel
On Sun, May 29, 2011 at 10:48 AM, ankit sambyal ankitsamb...@gmail.comwrote:
@sravanreddy: it won't work. Try 3,91,9 and 90,1,8,2,5 . correct
me if i m wrong.
Thanks,
Ankit Sambyal
On Sat, May 28, 2011 at 9:16 PM, ross jagadish1...@gmail.com wrote:
@sravanreddy:
Hey, Nice Solution :) cool!
On May 29, 7:44 am, sravanreddy001 sravanreddy...@gmail.com wrote:
Maintain a pointer A_end = m-1;
doing a comparision something similar to merge sort
int i=0;j=0;
while (i m){
if (a[i] b[j])
i++;
else{
swap(a[A_end],b[j])
A_end --;
j++;
}
}
This runs in O(m) time and no extra space, also the sort order is not
guarenteed.
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To post to this group, send email to algogeeks@googlegroups.com.
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